This cheat sheet will list several common geometrical problems found in games, and how to solve them with vector math.

Complete list of basic vector operations

But first, a little review.

For this, I assume that you have a vector class readily available. This is mostly 2D-focused, but everything works the same for 3D, except for differences concerning vector product, which I will assume to return just a scalar in the 2D case, representing the “z” axis. Any case that only applies to 2D or 3D will be pointed out.

Strictly speaking, a point is not a vector – but a vector can be used to represent the distance from the origin (0, 0) to the point, and so, it is perfectly reasonable to just use vectors to represent positions as if they were points.

I expect the class to give you access to each of the components, and to the following operations (using C++ style notation, including operator overloading – but it should be easy to translate to any other language of your choice). If a given operation is not available, you can still do it manually, either by extending the class or creating a “VectorUtils” class. The examples below are usually for 2D vectors – but 3D is usually simply a matter of adding the z coordinate following the pattern of x and y.

Vector2f operator+(Vector2f vec): Returns the sum of the two vectors. (In a language without operator overloading, this will probably be called add(). Similarly for the next few ones.)a + b = Vector2f(a.x + b.x, a.y + b.y);

float squaredLength(): Returns the square of the length of the vector. Useful when you just want to compare two vectors to see which is longest, as this avoids computing square rootsa.squaredLength() = a.x * a.x + a.y * a.y;

float unit(): Returns a vector pointing on the same direction, but with a length of 1.a.unit() = a / a.length();

Simple cases – warming up

Case #01 – Distance between two points

You probably know that this is done with the Pythagorean theorem, but the vectorial way is simpler. Given two vectors a and b:

float distance = (a-b).length();

Case #02 – Alignment

Sometimes, you want to align an image by its center. Sometimes, by its top-left corner. Or sometimes, by its top-center point. More generally, you can do alignment using a vector whose two components go from 0 to 1 (or even beyond, if you’d like), giving you full control of alignment.

Case #03 – Parametric Line Equation

Two points define a line, but it can be tricky to do much with this definition. A better way to work with a line is its parametric equation: one point (“P0″) and a direction vector (“dir”).

Vector2f p0 = point1;
Vector2f dir = (point2 - point1).unit();

With this, you can, for example, get a point 10 units away by simply doing:

Vector2f p1 = p0 + dir * 10;

Case #04 – Midpoint and interpolation between points

Say you have vectors p0 and p1. The midpoint between them is simply (p0+p1)/2. More generally, the line segment defined by p0 and p1 can be generated by varying t between 0 and 1 in the following linear interpolation:

Vector2f p = (1-t) * p0 + t * p1;

At t = 0, you get p0; at t = 1, you get p1; at t = 0.5, you get the midpoint, etc.

Case #05 – Finding the normal of a line segment

You already know how to find the direction vector of a line segment (case #03). The normal vector is a 90 degree rotation of that, so just call turnLeft() or turnRight() on it!

Projections using the Dot Product

The dot product has the incredibly useful property of being able to compute the length of a vector’s projection along the axis of another. To do this, you need the vector that you’ll project (“a“) and a unit vector (so make sure that you call unit() on it first!) representing the direction (“dir“). The length is then simply a.dot(dir). For example, if you have a = (3, 4) and dir = (1, 0), then a.dot(dir) = 3, and you can tell that this is correct, because (1, 0) is the direction vector of the x axis. In fact, a.x is always equivalent to a.dot(Vector2f(1, 0)), and a.y is equivalent to a.dot(Vector2f(0, 1)).

Because the dot product between a and b is also defined as |a||b|cos(alpha) (where alpha is the angle between the two), the result will be 0 if the two vectors are perpendicular, positive if the angle between them is less than 90, and negative if greater. This can be used to tell if two vectors point in the same general direction.

If you multiply the result of that dot product by the direction vector itself, you get the vector projected along that axis – let’s call that “at” (t for tangent). If you now do a – at, you get the part of the vector that is perpendicular to the dir vector – let’s call that “an” (n for normal). at + an = a.

Case #06 – Determining direction closest to dir

Say that you have a list of directions represented as unit vectors, and you want to find which of them is the closest to dir. Simply find the largest dot product between dir and a vector in the list. Likewise, the smallest dot product will be the direction farthest away.

Case #07 – Determining if the angle between two vectors is less than alpha

Using the equation above, we know that the angle between two vectors a and b will be less than alpha if the dot product between their unit vectors is less than cosine of alpha.

Case #08 – Determining which side of a half-plane a point is on

Say that you have an arbitrary point in space, p0, and a direction (unit) vector, dir. Imagine that an infinite line goes by p0, perpendicular to dir, dividing the plane in two, the half-plane that dir points to, and the half-plane that it does not point to. How do I tell whether a point p is in the side pointed to by dir? Remember that dot product is positive when the angle between vectors is less than 90 degrees, so just project and check against that:

Case #10 – Checking/forcing a point inside a convex polygon

A convex polygon can be defined to be the intersection of several half-planes, one for each edge of the polygon. Their p0 is either vertex of the edge, and their dir is the edge’s inner-facing normal vector (e.g., if you wind clockwise, that’d be the turnRight() normal). A point is inside the polygon if and only if it’s inside all the half-planes. Likewise, you can force it to be inside the polygon (by moving to the closest edge) by applying the makeInsideHalfPlane algorithm with every half-plane. [ops, this actually only works if all angles are >= 90 degrees]

Case #11 – Reflecting a vector with a given normal

Pong-like game. Ball hits a sloped wall. You know the ball’s velocity vector and the wall’s normal vector (see case #05). How do you reflect it realistically? Simple! Just reflect the ball’s normal velocity, and preserve its tangential velocity.

For more realism, you can multiply velT and velN by constants representing friction and restitution, respectively.

Case #12 – Cancelling movement along an axis

Sometimes, you want to restrict movement in a given axis. The idea is the same as above: decompose in a normal and tangential speed, and just keep tangential speed. This can be useful, for example, if the character is following a rail.

Rotations

Case #13 – Rotating a point around a pivot

If used to represent a point in space, the rotate() method will rotate that point around the origin. That might be interesting, but is limiting. Rotating around an arbitrary pivot vector is simple and much more useful – simply subtract the pivot from it, as if translating so the origin IS the pivot, then rotate, then add the pivot back:

Case #14 – Determining which direction to turn towards

Say that you have a character that wants to rotate to face an enemy. He knows his direction, and the direction that he should be facing to be looking straight at the enemy. But should he turn left or right? The cross product provides a simple answer: curDir.cross(targetDir) will return positive if you should turn left, and negative if you should turn right (and 0 if you’re either facing it already, or 180 degrees from it).

Other Geometric Cases

Here are a few other useful cases that aren’t that heavily vector-based, but useful:

Case #15 – Isometric world to screen coordinates

Isometric game. You know where the (0, 0) of world is on the screen (let’s call that point origin and represent it with a vector), but how do you know where a given world (x, y) is on the screen? First, you need two vectors determining the coordinate base, a new x and y axes. For a typical isometric game, they can be bx = Vector2f(2, 1) and by = Vector2f(-2, 1) – They don’t necessarily have to be unit vectors. From now, it’s straightforward:

Case #16 – Isometric screen to world coordinates

Same case, but now you want to know which tile the mouse is over. This is more complicated. Since we know that (x’, y’) = (x * bx.x + y * by.x, x * bx.y + y * by.y) + origin, we can first subtract origin, and then solve the linear equation. Using Cramer’s Rule, except that we’ll be a little clever and use our 2D cross-product (see definition at the beginning of the article) to simplify things:

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About

I'm Rodrigo Monteiro, and I have been programming video games since 1997, ranging from Klik & Play to C++ and Java. I'm the creator of the Aegisub subtitling software (together with Niels Hansen), and I like joining game programming challenges such as the Allegro SpeedHack and the Global Game Jam.

I currently work as a professional video game programmer at Bossa Studios, in London, United Kingdom.